Question a:
f(6) = (6)² + 2(6) + 8
f(6) = 36 + 12 + 8
f(6) = 56
Question b:
f(x+4) = (x+4)² + 2(x+4) + 8
f(x+4) = x² + 8x + 16 + 2x + 8 + 8
f(x+4) = x² + 10x + 32
Question c:
f(-x) = (-x)² + 2(-x) + 8
f(-x) = x - 2x + 8
Answer:
im not gonna lie cat i have no clue.
Step-by-step explanation:
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
